Some time ago the following rather easy problem appeared in an online publication called "Problems in Elementary NT" by Hojoo Lee:

Prove that there are infinitely many positive integers $a$, $b$, $c$ that are consecutive terms of an arithmetic progression and also satisfy the condition that $ab+1$, $bc+1$, $ca+1$ are all perfect squares.

This can be done using Pell's equation. What is interesting however is that the following result for four numbers apparently holds:

Claim. There are no positive integers $a$, $b$, $c$, $d$ that are consecutive terms of an arithmetic progression and also satisfy the condition that $ab+1$, $ac+1$, $ad+1$, $bc+1$, $bd+1$, $cd+1$ are all perfect squares.

5 Answers
5

Starting from the equations in my previous answer, we get, by multiplying them in pairs,
$$(x-y)x(x+y)(x+2y) + (x-y)x + (x+y)(x+2y) + 1 = (z_1 z_6)^2\,,$$
$$(x-y)x(x+y)(x+2y) + (x-y)(x+y) + x(x+2y) + 1 = (z_2 z_5)^2\,,$$
$$(x-y)x(x+y)(x+2y) + (x-y)(x+2y) + x(x+y) + 1 = (z_3 z_4)^2\,.$$
Write $u = z_1 z_6$, $v = z_2 z_5$, $w = z_3 z_4$ and take differences to obtain
$$3 y^2 = u^2 - v^2 \qquad\text{and}\qquad y^2 = v^2 - w^2\,.$$
The variety $C$ in ${\mathbb P}^3$ described by these two equations is a smooth curve of genus 1 whose Jacobian elliptic curve is 24a1 in the Cremona database; this elliptic curve has rank zero and a torsion group of order 8. This implies that $C$ has exactly 8 rational points; up to signs they are given by $(u:v:w:y) = (1:1:1:0)$ and $(2:1:0:1)$. So $y = 0$ or $w = 0$. In the first case, we do not have an honest AP ($y$ is the difference). In the second case, we get the contradiction $abcd + ad + bc + 1 = 0$ ($a,b,c,d$ are supposed to be positive). So unless I have made a mistake somewhere, this proves that there are no such APs of length 4.

Addition: We can apply this to rational points on the surface. The case $y = 0$ gives a bunch of conics of the form
$$x^2 + 1 = z_1^2, \quad z_2 = \pm z_1, \quad \dots, \quad z_6 = \pm z_1\,;$$
the case $w = 0$ leads to $ad = -1$ or $bc = -1$. The second of these gives $ad + 1 < 0$, and the first gives $ac + 1 = (a^2 + 1)/3$, which cannot be a square. This shows that all the rational points are on the conics mentioned above; in particular, (weak) Bombieri-Lang holds for this surface.

@Michael: Very nice! Let $X$ denote the surface given by your earlier answer. Then if I understand correctly, you've given a dominant rational map $X\to C$, where $C$ has genus 1 and $#C(\mathbb{Q})=8$. It seems a little surprising that a surface of general type should map onto a genus 1 curve, although I guess the fact that $X$ is an intersection of quadrics gives it a chance. But did you have some a priori reason to suspect that $X$ admitted such a covering?
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Joe SilvermanFeb 12 '12 at 12:52

1

@Joe: I didn't have some good a priori reason to suspect that $X$ would map to a genus 1 curve, but the observation that the analogous problem for five terms admits such a map (to the "four squares in AP" curve) provided some motivation to try a little bit longer.
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Michael StollFeb 12 '12 at 13:13

Bravo! The identities look a bit less mysterious if the arithmetic progression is written symmetrically as $(x-3y,x-y,x+y,x+3y)$, when the products $(z_i z_{7-i})^2$ are invariant under $y \leftrightarrow -y$ (Then $u^2-v^2$ and $v^2-w^2$ are $12y^2$ and $4y^2$, which comes to the same curves over ${\bf Q}$.)
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Noam D. ElkiesFeb 12 '12 at 19:50

Can magma (or other software) detect the nontrivial $H^1$ of this surface which makes this analysis possible?
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Noam D. ElkiesFeb 13 '12 at 1:16

Hm. I guess you would want the $H^1$ of its desingularization, which might be a fairly complicated object. As damiano pointed out to me by email, the fact that there is a map to an elliptic curve must be related to the singularities being sufficiently bad (otherwise the surface would be simply connected).
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Michael StollFeb 15 '12 at 11:10

Already for three-term progressions it's somewhat surprising that
there are infinitely many solutions, because the usual probabilistic
guess for the expected number of solutions leads to a convergent sum:
a random number of size $M$ is a square with probability about
$M^{-1/2}$, so we're summing something like $1/(abc)$ over all three-term progressions
$(a,b,c)$, etc. To be sure such a guess cannot account for non-random
patterns arising from polynomial identities, but it does suggest that
past a certain point such identities will be the only source of solutions.

Now a mindless exhaustive search over progressions
$(x,x+y,x+2y)$ with $0 < x,y < 10^4$ finds only the first six examples
$$
(1,7),\phantom+
(4,26),\phantom+
(15,97),\phantom+
(56,362),\phantom+
(209,1351),\phantom+
(780,5042)
$$
of an infinite family associated with the solutions
$(2,1)$, $(7,4)$, $(26,15)$, $(97,56)$, $(362,209)$, $(1351,780)$, etc.
of the Pell equation $x^2-3y^2=1$. If it can be proved that these are
the only solutions then it will immediately follow that there are
no four-term arithmetic progressions with the same property.
But that seems like a very hard problem.

Here's the gp code; with a bound of $10^4$ it takes only
a few minutes. One can surely do better with a more intelligent
search procedure (e.g. start by finding all solutions of $ab+1=r^2$
by factoring $r^2-1$).

Isn't there a similar problem involving 1,3,8,120, that says something like there is no fifth member such that the product of any two is one less than a square? It might be also that there are finitely many quadruplets, with the nontrivial ones being not arithmetic progressions. Gerhard "Ask Me About Obscure Puzzles" Paseman, 2012.02.11
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Gerhard PasemanFeb 11 '12 at 20:37

1

I was reminded of this 1,3,8,120 puzzle too, because Martin Gardner wrote about it in one of his columns decades ago and reported that the nonexistence of a fifth positive integer was proved only with difficulty (including several 1000+ digit computations, back when that was impressive). But there are infinitely many such integer quadruplets. A few Google searches turned up vixra.org/pdf/0907.0024v1.pdf with citations going as far as Euler and Diophantus! But naturally none of these are four-term arithmetic progressions...
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Noam D. ElkiesFeb 11 '12 at 21:43

Thanks for the reply! This is essentially what I did to claim that I "checked it manually": I verified the correspondence between the solutions of the problem in question and $x^{2}-3y^{2}=1$ up to $10^{9}$ and then just assummed this to be true. I was hoping however to find some slick way of doing the four number case without refering to the three case :).
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Cosmin PohoataFeb 12 '12 at 5:10

You're welcome. You saw that by now M.Stoll has completely solved the 4-term problem, even over the rationals. For the 3-term problem, you write that you "checked manually" up to $10^9$; how did you do that? I used the factorization of $t^2-1$ in $bc+1=t^2$ to search up to $t = 10^8$ in a few hours, still finding only the Pell solutions up to $(a,b,c) = (7865521, 58709048, 109552575)$; so $t=10^9$ (if that's what you meant) is feasible too, though it would take a while (or a number of CPU's running in parallel) to finish − and it's certainly not "manual" in the usual sense of "by hand"...
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Noam D. ElkiesFeb 12 '12 at 19:56

According to Magma, the projective closure of the variety associated to the problem (given by the equations
$$x(x-y) + 1 = z_1^2, \quad (x+y)(x-y) + 1 = z_2^2, \quad (x+2y)(x-y) + 1 = z_3^2,$$
$$(x+y)x + 1 = z_4^2, \quad (x+2y)x + 1 = z_5^2, \quad (x+2y)(x+y) + 1 = z_6^2 \quad)$$
is an irreducible surface in ${\mathbb P}^8$ with 34 isolated singularities. Since it is a complete intersection of six quadrics, it should be of general type (and it has trivial rational points with $x = y = 0$ and slightly less trivial ones with $y = 0$, so reduction methods will not work), which makes it very likely that this is a hard question.

Added later: You may want to look at Question 73346 for an explanation by Noam Elkies of the reasoning behind this.

This suggests that indeed the rational points are likewise fully accounted for by finitely many curves. The question asked only for integer points, which could conceivably be more tractable, though I don't see how.
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Noam D. ElkiesFeb 11 '12 at 21:39

@Noam: The problem with three terms should give a K3 Surface. Are there any K3-related methods that might prove that (up to signs) all the integral points are on the curve that comes from the Pell equation?
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Michael StollFeb 11 '12 at 22:01

1

There are no solutions with five terms $a,b,c,d,e$ in arithmetic progression, since then $ab+1$, $ac+1$, $ad+1$, $ae+1$ would have to be four squares in arithmetic progression. Of course, this doesn't really help...
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Michael StollFeb 11 '12 at 22:23

@Michael: It might be an interesting K3 surface but I don't see how to use this to prove anything about its integral points (for which it's of "log-general type").
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Noam D. ElkiesFeb 12 '12 at 20:39

To complete this: From $d \ge 4bc$, it follows easily that $a,b,c,d$ cannot be an AP. Note also the following: Acdording to Lemma 13 in Dujella's paper linked to in his answer above, if $a,b,c$ is a Diophantine triple with $a < b < c$, then $c = a + b + 2 \sqrt{ab+1}$ or $c \ge 4ab$. If $a,b,c$ form an AP, then the second possibility cannot occur, and the first (together with the AP condition) then implies that $b$ is even and $y^2 - 3(b/2)^2 = 1$, where $y = b-a = c-b$. So it is indeed the case that all such triples come from this Pell equation. (This gives another proof.)
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Michael StollFeb 22 '12 at 19:21